hw20solutions - g ( x ) on all of R , we use that it is...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: March 26, 2010 Assignment 20 1. Show that if f and g are uniformly continuous on A R and if they are both bounded on A , then their product fg is uniformly continuous on A . Since f and g are both bounded on A , there exist constants M 1 and M 2 so that | f ( x ) | ≤ M 1 and | g ( x ) | ≤ M 2 on A . Fix ε > 0. Since f and g are uniformly continuous, there exist δ 1 2 > 0 so that | f ( x ) - f ( y ) | < ε/ 2 M 2 for any x,y with | x - y | < δ 1 and | g ( x ) - g ( y ) | < varepsilon/ 2 M 1 for any x,y with | x - y | < δ 2 . Then setting δ = min( δ 1 2 ), we see that for any x,y A with | x - y | < δ , | f ( x ) g ( x ) - f ( y ) g ( y ) | ≤ | f ( x ) g ( x ) - f ( y ) g ( x ) | + | f ( y ) g ( x ) - f ( y ) g ( y ) | M 2 | f ( x ) - f ( y ) | + M 2 | g ( x ) - g ( y ) | < ε as desired. 2. If f ( x ) := x and g ( x ) := sin x , show that both f and g are uniformly continuous on R , but that their product fg is not uniformly continuous on R . For any given ε > 0, one can simply choose δ = ε to show the uniform continuity of f ( x ) = x . For g ( x ), we know that sin is continuous. Thus, on the compact set [ - 2 π, 2 π ], g ( x ) is uniformly continuous. Hence, given ε > 0, there is a δ > 0 so that if | x - y | < δ and x,y [ - 2 π,π ], then | g ( x ) - g ( y ) | < ε . To get uniform continuity of
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: g ( x ) on all of R , we use that it is periodic. Thus, for any x,y R so that | x-y | &lt; (assuming without loss that &lt; 1) we can substract 2 n for some n Z so that x-2 n and y-2 n are in [-2 , 2 ]. Then, | sin x-sin y | = | sin( x-2 n )-sin( y-2 n ) | &lt; since | ( x-2 n )-( y-2 n ) | &lt; . To show nonuniform continuity of fg on R , we shall use that sin x/x 1 as x and Theorem 5.4.2. Indeed, let x n = 2 n and u n = 2 n + 1 n . Then x n-u n 0. x n sin x n = 0 0. But by the sine addition formula, u n sin u n = (2 n + 1 /n ) sin 1 /n . But 2 n sin 1 /n 2 . Thus for n suciently large, | fg ( x n )-fg ( u n ) | = (2 n + 1 /n ) sin 1 /n 1 / 2 . Hence by the Nonuniform Continuity Criterion, fg is not uniformly continous. 1...
View Full Document

This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.

Ask a homework question - tutors are online